calculate-I-0-pi-2-x-3-x-cos-2-xdx-and-J-0-pi-2-x-3-x-sin-2-xdx-cslculate-I-and-J-

Question Number 36415 by abdo.msup.com last updated on 01/Jun/18

c a l c u l a t e I = \int_{0}^{\frac{π}{2}} (x^{3} + x) {c o s}^{2} x d x a n d

J = \int_{0}^{\frac{π}{2}} (x^{3} + x) {s i n}^{2} x d x

c s l c u l a t e I a n d J .

Commented by abdo.msup.com last updated on 04/Jun/18

w e h a v e I + J = \int_{0}^{\frac{π}{2}} (x^{3} + x) d x

= {[\frac{x^{4}}{4} + \frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{4}}{4 . 2^{4}} + \frac{π^{2}}{2 . 2^{2}} = \frac{π^{4}}{64} + \frac{π^{2}}{8}

a l s o w e h a v e I - J = \int_{0}^{\frac{π}{2}} (x^{3} + x) c o s (2 x) d x

=_{2 x = t} \int_{0}^{π} {{(\frac{t}{2})}^{3} + \frac{t}{2}} c o s (t) \frac{d t}{2}

= \frac{1}{2} \int_{0}^{π} {\frac{t^{3}}{8} + \frac{t}{2}} c o s (t) d t

= \frac{1}{16} \int_{0}^{π} {t^{3} + 4 t) c o s (2 t) d t b y p a r t s

\int_{0}^{π} {t^{3} + 4 t) c o s (2 t) = {[\frac{1}{2} (t^{3} + 4 t) s i n (2 t)]}_{0}^{π}

- \frac{1}{2} \int_{0}^{π} {3 t^{2} + 4} s i n (2 t) d t

= - \frac{1}{2} {{[- \frac{1}{2} (3 t^{2} + 4) c o s (2 t)]}_{0}^{π}

- \int_{0}^{π} - \frac{1}{2} (6 t) c o s (2 t) d t

= \frac{1}{4} {(3 π^{2} + 4) - 4} - \frac{3}{2} \int_{0}^{π} t c o s (2 t) d t

= \frac{3}{4} π^{2} - \frac{3}{2} {{[\frac{1}{2} t s i n (2 t)]}_{0}^{π} - \int_{0}^{π} \frac{1}{2} s i n (2 t) d t}

= \frac{3 π^{2}}{4} + \frac{3}{4} \int_{0}^{π} s i n (2 t) d t

= \frac{3 π^{2}}{4} - \frac{3}{8} {[c o s (2 t)]}_{0}^{π} = \frac{3 π^{2}}{4} s o

I + J = \frac{π^{4}}{64} + \frac{π^{2}}{8} a n d I - J = \frac{3 π^{2}}{4} \Rightarrow

2 I = \frac{π^{4}}{64} + \frac{7 π^{2}}{8} \Rightarrow I = \frac{π^{4}}{128} + \frac{7 π^{2}}{16} a n d

2 J = \frac{π^{4}}{64} + \frac{π^{2}}{8} - \frac{3 π^{2}}{4} = \frac{π^{4}}{64} - \frac{5 π^{2}}{8} \Rightarrow

J = \frac{π^{4}}{128} - \frac{5 π^{2}}{16} .